If $f = \{(1,2), (2,3), (3,1)\}$,it is easy to see that $f$ is one-one and onto. Find the inverse function $f^{-1}$.

Consider $f: \{1, 2, 3\} \rightarrow \{a, b, c\}$ and $g: \{a, b, c\} \rightarrow \{\text{apple, ball, cat}\}$ defined as $f(1)=a, f(2)=b, f(3)=c$ and $g(a)=\text{apple}, g(b)=\text{ball}, g(c)=\text{cat}$. Show that $f, g$ and $g \circ f$ are invertible. Find $f^{-1}, g^{-1}$ and $(g \circ f)^{-1}$ and show that $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$.

If $f(x) = \exp(2x^3 + 3x^2 + 6x)$ and $g(x)$ is the inverse function of $f(x)$,then the value of $g'(e^{11})$ is -

Let $f: X \rightarrow Y$ be an invertible function. Show that $f$ has a unique inverse.
(Hint: Suppose $g_{1}$ and $g_{2}$ are two inverses of $f$. Then for all $y \in Y$,$f \circ g_{1}(y) = I_{Y}(y) = f \circ g_{2}(y)$. Use the one-one property of $f$.)

If $f(x) = \int\limits_0^x {\frac{1}{{\sqrt {1 + {t^3}} }}\,} dt$ and $h(x)$ is the inverse of $f(x)$,then the value of $\frac{{h''(x)}}{{{h^2}(x)}}$ is

If the functions $f$ and $g$ are defined by $f(x) = 3x - 4$ and $g(x) = 2 + 3x$ for $x \in R$,then $g^{-1}(f^{-1}(5))$ is equal to